Q.The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

Concept & Intuition

Kinetic energy depends only on the speed of an object, not its direction. The velocity vector tells us the direction and magnitude; we need to extract the magnitude (the speed) by taking the square root of the sum of squares of its components. Then we plug that speed into the scalar kinetic energy formula. A common mistake is to treat the vector’s components as if they contribute separately to kinetic energy — but energy is a scalar, so only the total speed matters.

Step-by-step solution

1. Recall the kinetic energy formula For a body of mass $m$ moving with speed $v$,

Here $K = 87$ J is given, and we need $m$.

2. Find the speed from the velocity vector Velocity $\vec{v} = 2\hat{i} + 3\hat{j} - 4\hat{k}$ m/s. Speed $v$ is the magnitude:

So $v^2 = 29$ m²/s².

3. Solve for mass Rearrange the kinetic energy formula:

Substitute:

4. Match with options The value 6 kg corresponds to option (D).

The problem gives you a velocity vector and a kinetic energy, and asks for mass. The key idea is that kinetic energy is a scalar — it doesn't care about which way the body is moving, only how fast. So you never need to worry about the individual components of $\vec{v}$ beyond finding the speed.

The formula for kinetic energy is $K = \frac12 m v^2$, where $v$ is the magnitude of velocity (the speed). If you know $K$ and $v$, you can solve for $m$.

1. Find the speed from the velocity vector. The velocity is $\vec{v} = 2\hat{i} + 3\hat{j} - 4\hat{k}$ m/s. The magnitude is:

2. Use the kinetic energy equation. Given $K = 87$ J, we have:

Since $(\sqrt{29})^2 = 29$, this becomes:

3. Solve for $m$. Multiply both sides by 2:

Then divide by 29:

When working with vectors, we often need to describe a direction without being concerned about the magnitude. This is precisely the purpose of a unit vector. A unit vector is a vector that has a magnitude of 1 and points in a specific direction. To find the unit vector in the direction of any given vector, say $\vec{V}$, we simply divide the vector $\vec{V}$ by its own magnitude, $|\vec{V}|$. This process normalizes the vector, scaling it down (or up) so that its length becomes exactly 1, while preserving its original direction.

In this problem, we are asked to find the unit vector along the sum of two vectors, $\vec{A}$ and $\vec{B}$. This means our first step must be to find the resultant vector $\vec{R} = \vec{A} + \vec{B}$. Once we have this resultant vector, we will calculate its magnitude and then divide $\vec{R}$ by $|\vec{R}|$ to obtain the desired unit vector.

Here's how we proceed:

1. Calculate the resultant vector $\vec{R} = \vec{A} + \vec{B}$:

   Vector addition is performed by adding the corresponding components (x, y, and z) of the individual vectors.

   Given:

   $\vec{A} = \hat{i} - 2\hat{j} + 3\hat{k}$

   $\vec{B} = 2\hat{i} - 4\hat{j} + 3\hat{k}$

   Adding them:

   $\vec{R} = (1\hat{i} + 2\hat{i}) + (-2\hat{j} - 4\hat{j}) + (3\hat{k} + 3\hat{k})$

   $\vec{R} = (1+2)\hat{i} + (-2-4)\hat{j} + (3+3)\hat{k}$

   $\vec{R} = 3\hat{i} - 6\hat{j} + 6\hat{k}$

   Watch out

   A common mistake is to forget the signs of the components during addition. Always pay close attention to whether a component is positive or negative.

2. Calculate the magnitude of the resultant vector $|\vec{R}|$:

   The magnitude of a vector $\vec{V} = V_x\hat{i} + V_y\hat{j} + V_z\hat{k}$ is given by the formula:

   $|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}$

   For our resultant vector $\vec{R} = 3\hat{i} - 6\hat{j} + 6\hat{k}$:

   $|\vec{R}| = \sqrt{(3)^2 + (-6)^2 + (6)^2}$

   $|\vec{R}| = \sqrt{9 + 36 + 36}$

   $|\vec{R}| = \sqrt{81}$

   $|\vec{R}| = 9$

3. Determine the unit vector along $\vec{R}$:

   The unit vector in the direction of $\vec{R}$, denoted as $\hat{R}$, is found by dividing the vector $\vec{R}$ by its magnitude $|\vec{R}|$.

   $\hat{R} = \frac{\vec{R}}{|\vec{R}|}$

   $\hat{R} = \frac{3\hat{i} - 6\hat{j} + 6\hat{k}}{9}$

   Now, distribute the division to each component:

   $\hat{R} = \frac{3}{9}\hat{i} - \frac{6}{9}\hat{j} + \frac{6}{9}\hat{k}$

   $\hat{R} = \frac{1}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$

   This can also be written as:

   $\hat{R} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3}$

4. Compare with the given options:

   (A) $3\hat{i} - 6\hat{j} + 6\hat{k}$ (This is $\vec{A}+\vec{B}$, not the unit vector)

   (B) $\dfrac{\hat{i} - 2\hat{j} + 2\hat{k}}{3}$ (This matches our calculated unit vector)

   (C) $\dfrac{\hat{i} + 2\hat{j} + 2\hat{k}}{7}$

   (D) $\dfrac{3\hat{i} + 4\hat{j} + 5\hat{k}}{3}$

   The calculated unit vector matches option (B).

The key idea here is that work done by a constant force (gravity) is simply force times displacement in the direction of the force. Since gravitational force is constant ($mg$), the work done in any time interval is proportional to the distance fallen in that interval. So the ratio of works reduces to the ratio of distances covered in the first two seconds and the next two seconds of free fall.

For a body starting from rest under gravity ($g = 10 \, \text{m/s}^2$ is standard for such problems), the distance fallen in time $t$ is $s = \frac{1}{2} g t^2$. This is a quadratic in $t$, so distances in successive equal time intervals follow the pattern $1:3:5:7\ldots$ — a classic result from kinematics.

Let’s work it out step by step.

1. Distance fallen in the first 2 seconds Using $s = \frac{1}{2} g t^2$ with $t = 2$ s and $g = 10 \, \text{m/s}^2$:

2. Distance fallen in the first 4 seconds (i.e., up to the end of the “next two seconds”)

3. Distance fallen in the next two seconds (from $t=2$ to $t=4$) This is simply the difference:

4. Work done by gravity Work = force $\times$ displacement = $mg \times$ distance fallen. Since $mg$ is constant, the ratio of works equals the ratio of distances:

The key idea is that electric flux $\Phi = \vec{E} \cdot \vec{A}$ — a scalar product. That means only the component of the electric field perpendicular to the surface (i.e., parallel to the area vector) contributes. If the field has components that lie in the plane of the surface, they produce zero flux.

Here, the area lies in the XZ plane. The area vector is always perpendicular to the surface, so for the XZ plane, the perpendicular direction is the Y-axis. Thus $\vec{A} = A\,\hat{j}$ (or $-\hat{j}$, but magnitude is what matters for flux magnitude).

Let’s work through it:

1. Identify the area vector.

   The area is $3\ \mathrm{m}^2$ and lies in the XZ plane. The normal to the XZ plane is the Y-axis. So we take $\vec{A} = 3\,\hat{j}\ \mathrm{m}^2$.

2. Write the electric field.

   $\vec{E} = 8\hat{i} + 13\hat{j}\ \mathrm{NC}^{-1}$.

3. Compute the dot product.

   $\Phi = \vec{E} \cdot \vec{A} = (8\hat{i} + 13\hat{j}) \cdot (3\hat{j})$.

   Since $\hat{i} \cdot \hat{j} = 0$, the $8\hat{i}$ term contributes nothing.

   So $\Phi = 13 \times 3 = 39\ \mathrm{N\,m^2/C}$.

4. Units.

   $1\ \mathrm{N\,m^2/C} = 1\ \mathrm{Wb}$ (weber), so the flux is $39\ \mathrm{Wb}$.

Concept & Intuition

A soap bubble has two surfaces — an inner and an outer layer — each with surface tension. When you blow a bubble, you do work against surface tension to increase the total surface area. That work is stored as surface energy. The key is: work done = surface tension × total increase in surface area. For a bubble of radius $R$, the total surface area (both sides) is $2 \times 4\pi R^2 = 8\pi R^2$. So the work $W$ is proportional to $R^2$. If you double the radius, the area (and thus the work) multiplies by $2^2 = 4$.

Step-by-step reasoning

1. Work done = surface energy gained When a soap bubble is formed, the work done $W$ equals the product of surface tension $S$ and the total surface area created. Since a bubble has two surfaces, the total area for radius $R$ is $A_R = 2 \times 4\pi R^2 = 8\pi R^2$. Hence:

2. For radius $2R$ The total surface area becomes:

3. Work for the larger bubble Let the work done be $W'$:

4. Compare the two Divide $W'$ by $W$:

So $W' = 4W$.

The process of dividing a large liquid drop into smaller ones increases the total surface area of the liquid. Liquid surfaces possess surface energy, which is directly proportional to their surface area. To increase the surface area, work must be done against the cohesive forces between the liquid molecules, which manifest as surface tension. This work done is stored as additional surface energy.

The fundamental concept here is that the work done ($W$) in changing the surface area of a liquid is given by:

We need to calculate the initial total surface area and the final total surface area to find $\Delta A$.

Here's how we approach the problem:

1. Determine the radius of the smaller droplets:

   The total volume of the liquid remains constant throughout the process. The original spherical drop has a radius $r$. Let the radius of each of the 8 smaller droplets be $r'$.

   The volume of the original drop is $V_{initial} = \frac{4}{3}\pi r^3$.

   The volume of one small droplet is $V_{droplet} = \frac{4}{3}\pi (r')^3$.

   Since the original drop is divided into 8 equal droplets, the total final volume is $V_{final} = 8 \times \frac{4}{3}\pi (r')^3$.

   Equating the initial and final volumes:

   $\frac{4}{3}\pi r^3 = 8 \times \frac{4}{3}\pi (r')^3$

   $r^3 = 8 (r')^3$

   Taking the cube root of both sides:

   $r = 2r'$

   This gives us the radius of each small droplet: $r' = \frac{r}{2}$.

2. Calculate the initial total surface area:

   The original drop is a single sphere of radius $r$.

   Its surface area is $A_{initial} = 4\pi r^2$.

3. Calculate the final total surface area:

   Each small droplet is a sphere of radius $r' = \frac{r}{2}$.

   The surface area of one small droplet is $A_{one\_droplet} = 4\pi (r')^2 = 4\pi \left(\frac{r}{2}\right)^2 = 4\pi \frac{r^2}{4} = \pi r^2$.

   Since there are 8 such droplets, the total final surface area is $A_{final} = 8 \times A_{one\_droplet} = 8 \times \pi r^2$.

4. Calculate the change in total surface area ($\Delta A$):

   The change in surface area is the final total surface area minus the initial total surface area.

   $\Delta A = A_{final} - A_{initial}$

   $\Delta A = 8\pi r^2 - 4\pi r^2$

   $\Delta A = 4\pi r^2$

5. Calculate the work done:

   Using the formula $W = S \times \Delta A$:

   $W = S \times (4\pi r^2)$

   $W = 4\pi r^2 S$

When a box is pulled on a rough horizontal surface, several forces act on it. To determine its acceleration, we need to understand how these forces interact and apply Newton's laws of motion. The key idea is to break down the problem into components: first, analyze the forces in the vertical direction to find the normal force, and then use that to calculate the friction force. Once we have all horizontal forces, we can find the net horizontal force and thus the acceleration.

Here's how we approach this problem:

1. Identify and Resolve Forces:

   The box experiences four main forces:

   • Weight ($mg$): Acting vertically downwards due to gravity.
   • Normal Force ($N$): Acting vertically upwards, exerted by the surface on the box.
   • Applied Force ($F$): Given as $40\, \mathrm{N}$ at an angle. The problem states the angle is $60^\circ$ from the vertical. This means the angle with the horizontal is $90^\circ - 60^\circ = 30^\circ$. We need to resolve this force into its horizontal ($F_x$) and vertical ($F_y$) components.
   • Kinetic Friction Force ($f_k$): Acting horizontally, opposite to the direction of motion, because the surface is rough and the box is moving.

   Let's calculate the components of the applied force $F$:

   • Horizontal component: $F_x = F \cos(30^\circ)$
   • Vertical component: $F_y = F \sin(30^\circ)$

   Given $F = 40\, \mathrm{N}$:

   $F_x = 40 \cos(30^\circ) = 40 \left(\frac{\sqrt{3}}{2}\right) = 20\sqrt{3}\, \mathrm{N}$

   $F_y = 40 \sin(30^\circ) = 40 \left(\frac{1}{2}\right) = 20\, \mathrm{N}$

   Watch out

   Always pay close attention to the angle given. If the angle is specified "from vertical", it's often easier to convert it to an angle "from horizontal" for standard component resolution. Here, $60^\circ$ from vertical means $30^\circ$ from horizontal.

2. Analyze Vertical Forces to Find Normal Force:

   The box is moving horizontally, so there is no acceleration in the vertical direction. This means the net force in the vertical direction is zero.

   The forces acting vertically are the normal force ($N$) upwards, the weight ($mg$) downwards, and the vertical component of the applied force ($F_y$) upwards.

   Using Newton's first law for vertical equilibrium:

   $N + F_y - mg = 0$

   $N = mg - F_y$

   Given $m = 10\, \mathrm{kg}$ and $g = 10\, \mathrm{m/s^2}$:

   $mg = 10\, \mathrm{kg} \times 10\, \mathrm{m/s^2} = 100\, \mathrm{N}$

   Substitute the values:

   $N = 100\, \mathrm{N} - 20\, \mathrm{N} = 80\, \mathrm{N}$

3. Calculate Kinetic Friction Force:

   The kinetic friction force is directly proportional to the normal force.

   The kinetic friction force $f_k$ is given by:

   $f_k = \mu_k N$

   where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force.

   Given $\mu_k = 0.25$ and $N = 80\, \mathrm{N}$:

   $f_k = 0.25 \times 80\, \mathrm{N} = 20\, \mathrm{N}$

4. Analyze Horizontal Forces to Find Acceleration:

   Now we apply Newton's second law in the horizontal direction. The net horizontal force causes the acceleration of the box.

   The horizontal forces are the horizontal component of the applied force ($F_x$) in the direction of motion and the kinetic friction force ($f_k$) opposing the motion.

   Newton's Second Law states that the net force $F_{net}$ acting on an object is equal to the product of its mass $m$ and acceleration $a$:

   $F_{net} = ma$

   The net horizontal force $F_{net,x}$ is:

   $F_{net,x} = F_x - f_k$

   Substitute the values:

   $F_{net,x} = 20\sqrt{3}\, \mathrm{N} - 20\, \mathrm{N}$

   Using $\sqrt{3} \approx 1.732$:

   $F_{net,x} = 20(1.732)\, \mathrm{N} - 20\, \mathrm{N}$

   $F_{net,x} = 34.64\, \mathrm{N} - 20\, \mathrm{N} = 14.64\, \mathrm{N}$

   Now, use Newton's second law to find the acceleration $a$:

   $a = \frac{F_{net,x}}{m}$

   $a = \frac{14.64\, \mathrm{N}}{10\, \mathrm{kg}} = 1.464\, \mathrm{m/s^2}$

   Rounding to two decimal places, the acceleration is $1.46\, \mathrm{m/s^2}$.

The calculated acceleration matches option (C).

The key idea here is that elastic potential energy in a stretched wire is the work done by the internal restoring force, which for a linear elastic material (obeying Hooke’s law) equals half the product of the force and the extension. But a more direct route uses stress and strain: since stress is proportional to strain, the energy per unit volume is $\frac{1}{2} \times \text{stress} \times \text{strain}$. Multiply by the volume to get the total stored energy. This avoids needing to compute the force explicitly.

Let’s work through it step by step.

1. Identify the given quantities and convert to SI units.

   • Length of wire: $L = 4 \, \text{m}$
   • Extension: $\Delta L = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m}$
   • Cross-sectional area: $A = 8 \, \text{mm}^2 = 8 \times 10^{-6} \, \text{m}^2$
   • Young’s modulus: $Y = 2.0 \times 10^{11} \, \text{N/m}^2$

2. Compute the strain.

   Strain is the fractional change in length:

3. Compute the stress using Young’s modulus. Young’s modulus relates stress and strain:

So:

4. Find the volume of the wire. Volume = area × length:

5. Apply the formula for elastic potential energy. For a material obeying Hooke’s law, the energy stored per unit volume is $\frac{1}{2} \times \text{stress} \times \text{strain}$. Therefore:

Substitute the values:

First multiply the stress and strain:

Then multiply by volume:

Finally, multiply by $\frac{1}{2}$:

Concept

Stretching a wire does work against interatomic forces, stored as elastic potential energy. For a linearly elastic material, the energy density is $\frac{1}{2}\times\text{stress}\times\text{strain}$; multiplying by volume gives the total stored energy.

Step-by-step solution

1. Given quantities

   • $L = 4\ \text{m}$
   • $\Delta L = 4\ \text{mm} = 4\times10^{-3}\ \text{m}$
   • $A = 8\ \text{mm}^2 = 8\times10^{-6}\ \text{m}^2$
   • $Y = 2.0\times10^{11}\ \text{N/m}^2$

2. Formula

3. Substitute

   $(\Delta L)^2 = (4\times10^{-3})^2 = 16\times10^{-6}\ \text{m}^2$

   Numerator: $Y\cdot A\cdot(\Delta L)^2 = (2.0\times10^{11})(8\times10^{-6})(16\times10^{-6}) = 25.6$

   Divide by $L=4$: $25.6/4 = 6.4$

   Multiply by $\tfrac{1}{2}$: $U = 3.2\ \text{J}$

4. Cross-check via force: the final tension is $F = \dfrac{YA\Delta L}{L} = \dfrac{(2.0\times10^{11})(8\times10^{-6})(4\times10^{-3})}{4} = 1600\ \text{N}$. Since the force rises linearly from 0 to $F$ as the wire stretches, the work done is the area under the force–extension graph: $U=\frac{1}{2}F\Delta L = \frac{1}{2}(1600)(4\times10^{-3}) = 3.2\ \text{J}$ — matching the direct formula.

Concept. The block starts at rest and finally stops, so its net kinetic-energy change over the slide is zero. Therefore the work done by gravity equals the work done against friction. Because friction varies with position, its work must be integrated.

Forces on the $30^\circ$ incline. Along the incline gravity pulls with $mg\sin 30^\circ$. The friction force is

Energy balance over distance $L$ (both endpoints at rest):

Solve for $L$ (cancel the common $\tfrac{mg}{2}L$, with $L\neq 0$):